1. ## Isomorphism

Define for each $\lambda\in\mathbb{R}$ a natural action on $T_{\alpha}$ by rotation:
$
\tau(\lambda)(f)(t) = \left\{
\begin{array}{lr}
f(t+\lambda) & : t+\lambda \in [0,1]\\
\alpha^n(f(t+\lambda-n)) & : t+\lambda \in [n,n+1],n\in\mathbb{Z}
\end{array}
\right.
$

where $\alpha$ is an automorphism of the $C^*$-algebra A. Prove that the one parameter family $\lambda\mapsto \tau(\lambda)$ gives rise to an isomorphism

$
T_{\alpha}\times_{\alpha}\mathbb{R}\cong (A\times_{\alpha}\mathbb{Z})\otimes \mathbb{K}(L^{2}(S^1))
$

where $T_{\alpha}:=\{f:[0,1]\rightarrow A : f(1)=\alpha(f(0))\}$

2. Originally Posted by Mauritzvdworm
Define for each $\lambda\in\mathbb{R}$ a natural action on $T_{\alpha}$ by rotation:
$
\tau(\lambda)(f)(t) = \left\{
\begin{array}{lr}
f(t+\lambda) & : t+\lambda \in [0,1]\\
\alpha(f(t+\lambda-n)) & : t+\lambda \in [n,n+1],n\in\mathbb{Z}
\end{array}
\right.
$

where $\alpha$ is an automorphism of the $C^*$-algebra A. Prove that the one parameter family $\lambda\mapsto \tau(\lambda)$ gives rise to an isomorphism

$
T_{\alpha}\times_{\alpha}\mathbb{R}\cong (A\times_{\alpha}\mathbb{Z})\otimes \mathbb{K}(L^{2}(S^1))
$

where $T_{\alpha}:=\{f:[0,1]\rightarrow A : f(1)=\alpha(f(0))\}$
I can't help with this problem, which probably requires a fairly extensive knowledge of semidirect products (and in any case I am about to go away for a week). But the problem as stated has two obvious mistakes.

First, the definition of $\tau(\lambda)$ is wrong. It does not define an automorphism group. In fact, the definition of $\tau(\lambda)(f)(t)$ does not even define a continuous function of t at integer values of $t+\lambda$. The correct definition should presumably be $\tau(\lambda)(f)(t) = \alpha^n(f(t+\lambda-n))$ when $t+\lambda\in[n,n+1]$.

Second, the semidirect product $T_\alpha\rtimes_\alpha\mathbb{R}$ does not make sense. It should be $T_\alpha\rtimes_\lambda\mathbb{R}.$